Problem: $\vec w = (10,-15)$ $-\dfrac45\vec w= ($
Explanation: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $-\dfrac45 \vec{w}$ : $\begin{aligned} {-\dfrac45}\vec w = {-\dfrac45} \cdot (10,-15) &= \left({-\dfrac45} \cdot 10, {-\dfrac45} \cdot (-15)\right) \\\\ &= (-8,12) \end{aligned}$ The answer is $ (-8,12) $.